The expression you have provided is true because sin 2x = 2*sin x*cos x and cos 2x = (cos x)^2 - (sin x)^2

Let start with (cos x + sin x)/(cos x - sin x)

multiply the numerator and denominator by (cos x + sin x)

=> (cos x +...

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The expression you have provided is true because sin 2x = 2*sin x*cos x and cos 2x = (cos x)^2 - (sin x)^2

Let start with (cos x + sin x)/(cos x - sin x)

multiply the numerator and denominator by (cos x + sin x)

=> (cos x + sin x)(cos x + sin x)/(cos x - sin x)(cos x + sin x)

=> (cos x + sin x)^2/(cos x)^2 - (sin x)^2

=> [(cos x)^2 + 2*sin x*cos x + (sin x)^2]/cos 2x

=> (1 + sin 2x)/cos 2x

**This proves that (cos x + sin x)/(cos x - sin x) = (1 + sin 2x)/cos 2x**